If O is the circumcentre and O' is the orthoc | vedclass

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(B) Let the origin be at the circumcentre $O$. Then the position vectors of $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| = |\vec{b}|

Vedclass · Accepted Answer

$2\,\overrightarrow{O'O}$

Vedclass · Answer

(B) Let the origin be at the circumcentre $O$. Then the position vectors of $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| = |\vec{b}| = |\vec{c}| = R$.The position vector of the orthocentre $O'$ is given by $\vec{o'} = \vec{a} + \vec{b} + \vec{c}$.We need to calculate $\overrightarrow{O'A} + \overrightarrow{O'B} + \overrightarrow{O'C}$.Using the definition of position vectors relative to the origin $O$:$\overrightarrow{O'A} = \vec{a} - \vec{o'} = \vec{a} - (\vec{a} + \vec{b} + \vec{c}) = -(\vec{b} + \vec{c})$$\overrightarrow{O'B} = \vec{b} - \vec{o'} = \vec{b} - (\vec{a} + \vec{b} + \vec{c}) = -(\vec{a} + \vec{c})$$\overrightarrow{O'C} = \vec{c} - \vec{o'} = \vec{c} - (\vec{a} + \vec{b} + \vec{c}) = -(\vec{a} + \vec{b})$Summing these up:$\overrightarrow{O'A} + \overrightarrow{O'B} + \overrightarrow{O'C} = -(\vec{b} + \vec{c} + \vec{a} + \vec{c} + \vec{a} + \vec{b}) = -2(\vec{a} + \vec{b} + \vec{c}) = -2\vec{o'}$.Since $\vec{o'}$ is the vector $\overrightarrow{OO'}$,we have $\overrightarrow{O'A} + \overrightarrow{O'B} + \overrightarrow{O'C} = -2\overrightarrow{OO'} = 2\overrightarrow{O'O}$.

If $O$ is the circumcentre and $O'$ is the orthocentre of the triangle $ABC$,then $\overrightarrow{O'A} + \overrightarrow{O'B} + \overrightarrow{O'C} = $

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